Prove that the curves $x=y^{2}$ and $xy=k$ cut at right angles if $8k^{2}=1$.

(N/A) The equations of the given curves are $x=y^{2}$ and $xy=k$.
Substituting $x=y^{2}$ into $xy=k$,we get:
$y^{2} \cdot y = k \Rightarrow y^{3} = k \Rightarrow y = k^{1/3}$.
Therefore,$x = (k^{1/3})^{2} = k^{2/3}$.
Thus,the point of intersection is $(k^{2/3}, k^{1/3})$.
Differentiating $x=y^{2}$ with respect to $x$,we have:
$1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.
The slope of the tangent $(m_{1})$ to $x=y^{2}$ at $(k^{2/3}, k^{1/3})$ is $m_{1} = \frac{1}{2k^{1/3}}$.
Differentiating $xy=k$ with respect to $x$,we have:
$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
The slope of the tangent $(m_{2})$ to $xy=k$ at $(k^{2/3}, k^{1/3})$ is $m_{2} = -\frac{k^{1/3}}{k^{2/3}} = -\frac{1}{k^{1/3}}$.
Two curves intersect at right angles if the product of their slopes is $-1$:
$m_{1} \cdot m_{2} = -1$
$\left(\frac{1}{2k^{1/3}}\right) \cdot \left(-\frac{1}{k^{1/3}}\right) = -1$
$-\frac{1}{2k^{2/3}} = -1$
$2k^{2/3} = 1$.
Cubing both sides,we get:
$(2k^{2/3})^{3} = 1^{3} \Rightarrow 8k^{2} = 1$.
Hence,the curves cut at right angles if $8k^{2}=1$.